The first step is to rewrite the problem as follows.

\(\displaystyle \left(a + 3 b\right)^{2}= \left(a + 3*b\right) \cdot \left(a + 3*b\right)\)

Now we multiply the first parts of the first and second expression together.

\(\displaystyle a\cdota=a^{2}\)

Now we multiply the first term  of the first expression with the second term of the second expression.

\(\displaystyle a\cdot3 b=3 a b\)

Now we multiply the second term of the first expression with the first term of the second expression.

\(\displaystyle a\cdot3 b=3 a b\)

Now we multiply the last terms of each expression together.

\(\displaystyle 3 b\cdot3 b=9 b^{2}\)

Now we add all these results together, and we get.

\(\displaystyle a^{2} + 6 a b + 9 b^{2}\)

The first step is to rewrite the problem as follows.

\(\displaystyle \left(a + 12 b\right)^{2}= \left(a + 12*b\right) \cdot \left(a + 12*b\right)\)

Now we multiply the first parts of the first and second expression together.

\(\displaystyle a\cdota=a^{2}\)

Now we multiply the first term  of the first expression with the second term of the second expression.

\(\displaystyle a\cdot12 b=12 a b\)

Now we multiply the second term of the first expression with the first term of the second expression.

\(\displaystyle a\cdot12 b=12 a b\)

Now we multiply the last terms of each expression together.

\(\displaystyle 12 b\cdot12 b=144 b^{2}\)

Now we add all these results together, and we get.

\(\displaystyle a^{2} + 24 a b + 144 b^{2}\)

The first step is to rewrite the problem as follows.

\(\displaystyle \left(a + 18 b\right)^{2} = \left( a + 18*b \right) \cdot \left( a + 18*b \right)\)

Now we multiply the first parts of the first and second expression together.

\(\displaystyle a \cdot a = a^{2}\)

Now we multiply the first term  of the first expression with the second term of the second expression.

\(\displaystyle a \cdot 18 b = 18 a b\)

Now we multiply the second term of the first expression with the first term of the second expression.

\(\displaystyle a \cdot 18 b = 18 a b\)

Now we multiply the last terms of each expression together.

\(\displaystyle 18 b \cdot 18 b = 324 b^{2}\)

Now we add all these results together, and we get.

\(\displaystyle a^{2} + 36 a b + 324 b^{2}\)

In order to do this, we need to recall the formula for Pascal's Triangle.

\(\displaystyle \sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}\)

Since the exponent in the question is 12 we can replace \(\displaystyle \uptext{n}\), with 12 .

Now our equation looks like

\(\displaystyle \sum_{i=0}^{12} x^{- i + 12} y^{i} {\binom{12}{i}}\)

Now we compute the sum, term by term.

Term 1 : \(\displaystyle \binom{ 12 }{ 0 } x ^{\left( 12 - 0 \right)} y ^{ 0 }= x^{12}\)

Term 2 : \(\displaystyle \binom{ 12 }{ 1 } x ^{\left( 12 - 1 \right)} y ^{ 1 }= 12 x^{11} y\)

Term 3 : \(\displaystyle \binom{ 12 }{ 2 } x ^{\left( 12 - 2 \right)} y ^{ 2 }= 66 x^{10} y^{2}\)

Term 4 : \(\displaystyle \binom{ 12 }{ 3 } x ^{\left( 12 - 3 \right)} y ^{ 3 }= 220 x^{9} y^{3}\)

Term 5 : \(\displaystyle \binom{ 12 }{ 4 } x ^{\left( 12 - 4 \right)} y ^{ 4 }= 495 x^{8} y^{4}\)

Term 6 : \(\displaystyle \binom{ 12 }{ 5 } x ^{\left( 12 - 5 \right)} y ^{ 5 }= 792 x^{7} y^{5}\)

Term 7 : \(\displaystyle \binom{ 12 }{ 6 } x ^{\left( 12 - 6 \right)} y ^{ 6 }= 924 x^{6} y^{6}\)

Term 8 : \(\displaystyle \binom{ 12 }{ 7 } x ^{\left( 12 - 7 \right)} y ^{ 7 }= 792 x^{5} y^{7}\)

Term 9 : \(\displaystyle \binom{ 12 }{ 8 } x ^{\left( 12 - 8 \right)} y ^{ 8 }= 495 x^{4} y^{8}\)

Term 10 : \(\displaystyle \binom{ 12 }{ 9 } x ^{\left( 12 - 9 \right)} y ^{ 9 }= 220 x^{3} y^{9}\)

Term 11 : \(\displaystyle \binom{ 12 }{ 10 } x ^{\left( 12 - 10 \right)} y ^{ 10 }= 66 x^{2} y^{10}\)

Term 12 : \(\displaystyle \binom{ 12 }{ 11 } x ^{\left( 12 - 11 \right)} y ^{ 11 }= 12 x y^{11}\)

Term 13 : \(\displaystyle \binom{ 12 }{ 12 } x ^{\left( 12 - 12 \right)} y ^{ 12 }= y^{12}\)

Now we combine the expressions and we get

\(\displaystyle \\x^{12} + 12 x^{11} y + 66 x^{10} y^{2} + 220 x^{9} y^{3} + 495 x^{8} y^{4} + 792 x^{7} y^{5} + 924 x^{6} y^{6} + 792 x^{5} y^{7} + 495 x^{4} y^{8} + 220 x^{3} y^{9} + 66 x^{2} y^{10} + 12 x y^{11} + y^{12}\)

In order to do this, we need to recall the formula for Pascal's Triangle.

\(\displaystyle \sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}\)

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of \(\displaystyle \uptext{y}\), and the exponent of the original equation.

In this case \(\displaystyle n = 17\) and \(\displaystyle i = 9\)

Now we compute the following

\(\displaystyle {\binom{n}{i}} = \binom{ 17 }{ 9 } = 24310\)

In order to do this, we need to recall the formula for Pascal's Triangle.

\(\displaystyle \sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}\)

Since the exponent in the question is 18 we can replace \(\displaystyle \uptext{n}\), with 18 .

Now our equation looks like

\(\displaystyle \sum_{i=0}^{18} x^{- i + 18} y^{i} {\binom{18}{i}}\)

Now we compute the sum, term by term.

Term 1 : \(\displaystyle \binom{ 18 }{ 0 } x ^{\left( 18 - 0 \right)} y ^{ 0 }= x^{18}\)

Term 2 : \(\displaystyle \binom{ 18 }{ 1 } x ^{\left( 18 - 1 \right)} y ^{ 1 }= 18 x^{17} y\)

Term 3 : \(\displaystyle \binom{ 18 }{ 2 } x ^{\left( 18 - 2 \right)} y ^{ 2 }= 153 x^{16} y^{2}\)

Term 4 : \(\displaystyle \binom{ 18 }{ 3 } x ^{\left( 18 - 3 \right)} y ^{ 3 }= 816 x^{15} y^{3}\)

Term 5 : \(\displaystyle \binom{ 18 }{ 4 } x ^{\left( 18 - 4 \right)} y ^{ 4 }= 3060 x^{14} y^{4}\)

Term 6 : \(\displaystyle \binom{ 18 }{ 5 } x ^{\left( 18 - 5 \right)} y ^{ 5 }= 8568 x^{13} y^{5}\)

Term 7 : \(\displaystyle \binom{ 18 }{ 6 } x ^{\left( 18 - 6 \right)} y ^{ 6 }= 18564 x^{12} y^{6}\)

Term 8 : \(\displaystyle \binom{ 18 }{ 7 } x ^{\left( 18 - 7 \right)} y ^{ 7 }= 31824 x^{11} y^{7}\)

Term 9 : \(\displaystyle \binom{ 18 }{ 8 } x ^{\left( 18 - 8 \right)} y ^{ 8 }= 43758 x^{10} y^{8}\)

Term 10 : \(\displaystyle \binom{ 18 }{ 9 } x ^{\left( 18 - 9 \right)} y ^{ 9 }= 48620 x^{9} y^{9}\)

Term 11 : \(\displaystyle \binom{ 18 }{ 10 } x ^{\left( 18 - 10 \right)} y ^{ 10 }= 43758 x^{8} y^{10}\)

Term 12 : \(\displaystyle \binom{ 18 }{ 11 } x ^{\left( 18 - 11 \right)} y ^{ 11 }= 31824 x^{7} y^{11}\)

Term 13 : \(\displaystyle \binom{ 18 }{ 12 } x ^{\left( 18 - 12 \right)} y ^{ 12 }= 18564 x^{6} y^{12}\)

Term 14 : \(\displaystyle \binom{ 18 }{ 13 } x ^{\left( 18 - 13 \right)} y ^{ 13 }= 8568 x^{5} y^{13}\)

Term 15 : \(\displaystyle \binom{ 18 }{ 14 } x ^{\left( 18 - 14 \right)} y ^{ 14 }= 3060 x^{4} y^{14}\)

Term 16 : \(\displaystyle \binom{ 18 }{ 15 } x ^{\left( 18 - 15 \right)} y ^{ 15 }= 816 x^{3} y^{15}\)

Term 17 : \(\displaystyle \binom{ 18 }{ 16 } x ^{\left( 18 - 16 \right)} y ^{ 16 }= 153 x^{2} y^{16}\)

Term 18 : \(\displaystyle \binom{ 18 }{ 17 } x ^{\left( 18 - 17 \right)} y ^{ 17 }= 18 x y^{17}\)

Term 19 : \(\displaystyle \binom{ 18 }{ 18 } x ^{\left( 18 - 18 \right)} y ^{ 18 }= y^{18}\)

Now we combine the expressions and we get

\(\displaystyle \\x^{18} + 18 x^{17} y + 153 x^{16} y^{2} + 816 x^{15} y^{3} + 3060 x^{14} y^{4} + 8568 x^{13} y^{5} + 18564 x^{12} y^{6} + 31824 x^{11} y^{7} + 43758 x^{10} y^{8} + 48620 x^{9} y^{9} + 43758 x^{8} y^{10} + 31824 x^{7} y^{11} + 18564 x^{6} y^{12} + 8568 x^{5} y^{13} + 3060 x^{4} y^{14} + 816 x^{3} y^{15} + 153 x^{2} y^{16} + 18 x y^{17} + y^{18}\)

In order to do this, we need to recall the formula for Pascal's Triangle.

\(\displaystyle \sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}\)

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of \(\displaystyle {y}\), and the exponent of the original equation.

In this case \(\displaystyle n = 13\) and \(\displaystyle i = 7\)

Now we compute the following

\(\displaystyle {\binom{n}{i}} = \binom{ 13 }{ 7 } = 1716\)

In order to do this, we need to recall the formula for Pascal's Triangle.

\(\displaystyle \sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}\)

Since the exponent in the question is 8 we can replace \(\displaystyle \uptext{n}\), with 8 .

Now our equation looks like

\(\displaystyle \sum_{i=0}^{8} x^{- i + 8} y^{i} {\binom{8}{i}}\)

Now we compute the sum, term by term.

Term 1 : \(\displaystyle \binom{ 8 }{ 0 } x ^{\left( 8 - 0 \right)} y ^{ 0 }= x^{8}\)

Term 2 : \(\displaystyle \binom{ 8 }{ 1 } x ^{\left( 8 - 1 \right)} y ^{ 1 }= 8 x^{7} y\)

Term 3 : \(\displaystyle \binom{ 8 }{ 2 } x ^{\left( 8 - 2 \right)} y ^{ 2 }= 28 x^{6} y^{2}\)

Term 4 : \(\displaystyle \binom{ 8 }{ 3 } x ^{\left( 8 - 3 \right)} y ^{ 3 }= 56 x^{5} y^{3}\)

Term 5 : \(\displaystyle \binom{ 8 }{ 4 } x ^{\left( 8 - 4 \right)} y ^{ 4 }= 70 x^{4} y^{4}\)

Term 6 : \(\displaystyle \binom{ 8 }{ 5 } x ^{\left( 8 - 5 \right)} y ^{ 5 }= 56 x^{3} y^{5}\)

Term 7 : \(\displaystyle \binom{ 8 }{ 6 } x ^{\left( 8 - 6 \right)} y ^{ 6 }= 28 x^{2} y^{6}\)

Term 8 : \(\displaystyle \binom{ 8 }{ 7 } x ^{\left( 8 - 7 \right)} y ^{ 7 }= 8 x y^{7}\)

Term 9 : \(\displaystyle \binom{ 8 }{ 8 } x ^{\left( 8 - 8 \right)} y ^{ 8 }= y^{8}\)

Now we combine the expressions and we get

\(\displaystyle x^{8} + 8 x^{7} y + 28 x^{6} y^{2} + 56 x^{5} y^{3} + 70 x^{4} y^{4} + 56 x^{3} y^{5} + 28 x^{2} y^{6} + 8 x y^{7} + y^{8}\)

In order to do this, we need to recall the formula for Pascal's Triangle.

\(\displaystyle \sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}\)

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of \(\displaystyle \uptext{y}\), and the exponent of the original equation.

In this case \(\displaystyle n = 15\) and \(\displaystyle i = 6\)

Now we compute the following

\(\displaystyle {\binom{n}{i}} = \binom{ 15 }{ 6 } = 5005\)

In order to do this, we need to recall the formula for Pascal's Triangle.

\(\displaystyle \sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}\)

Since the exponent in the question is 7 we can replace \(\displaystyle \uptext{n}\), with 7 .

Now our equation looks like

\(\displaystyle \sum_{i=0}^{7} x^{- i + 7} y^{i} {\binom{7}{i}}\)

Now we compute the sum, term by term.

Term 1 : \(\displaystyle \binom{ 7 }{ 0 } x ^{\left( 7 - 0 \right)} y ^{ 0 }= x^{7}\)

Term 2 : \(\displaystyle \binom{ 7 }{ 1 } x ^{\left( 7 - 1 \right)} y ^{ 1 }= 7 x^{6} y\)

Term 3 : \(\displaystyle \binom{ 7 }{ 2 } x ^{\left( 7 - 2 \right)} y ^{ 2 }= 21 x^{5} y^{2}\)

Term 4 : \(\displaystyle \binom{ 7 }{ 3 } x ^{\left( 7 - 3 \right)} y ^{ 3 }= 35 x^{4} y^{3}\)

Term 5 : \(\displaystyle \binom{ 7 }{ 4 } x ^{\left( 7 - 4 \right)} y ^{ 4 }= 35 x^{3} y^{4}\)

Term 6 : \(\displaystyle \binom{ 7 }{ 5 } x ^{\left( 7 - 5 \right)} y ^{ 5 }= 21 x^{2} y^{5}\)

Term 7 : \(\displaystyle \binom{ 7 }{ 6 } x ^{\left( 7 - 6 \right)} y ^{ 6 }= 7 x y^{6}\)

Term 8 : \(\displaystyle \binom{ 7 }{ 7 } x ^{\left( 7 - 7 \right)} y ^{ 7 }= y^{7}\)

Now we combine the expressions and we get

\(\displaystyle x^{7} + 7 x^{6} y + 21 x^{5} y^{2} + 35 x^{4} y^{3} + 35 x^{3} y^{4} + 21 x^{2} y^{5} + 7 x y^{6} + y^{7}\)